Problem: Here's a parameterization of a plane: $\vec{v}(x, y) = (-5y, 2x + y, 4x - y)$ What vectors are normal to the plane $\vec{v}$ ? Choose 2 answers: Choose 2 answers: (Choice A) A $(6, 20, -10)$ (Choice B) B $(-6, -20, 10)$ (Choice C) C $(-20, -10, -6)$ (Choice D) D $(20, 10, 6)$
The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. We can take multiply the result by $-1$ to find another normal vector. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ 0 & 2 & 4 \\ \\ -5 & 1 & -1 \end{pmatrix} \\ \\ &= -6 \hat{\imath} - 20 \hat{\jmath} + 10 \hat{k} \end{aligned}$ We want two normal vectors. Because the negative of a normal vector is also normal to the surface, we can take the negative of what we just calculated as a second normal vector to the plane $\vec{v}$. Therefore, two vectors normal to $\vec{v}$ are $(-6, -20, 10)$ and $(6, 20, -10)$.